A ball of mass 800g moving horizontally with a speed of 5m/s hits a vertical wall and rebounds with the same speed. The impulse experienced by the ball is?

Answer Details

The impulse experienced by the ball can be calculated using the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. In this case, the momentum of the ball before the collision is: p1 = m * v1 where m is the mass of the ball and v1 is its velocity before the collision. Substituting the values given in the problem, we get: p1 = 0.8 kg * 5 m/s = 4 kg m/s After the collision, the ball rebounds with the same speed but in the opposite direction, so its velocity after the collision is: v2 = -5 m/s The momentum of the ball after the collision is: p2 = m * v2 Substituting the values, we get: p2 = 0.8 kg * (-5 m/s) = -4 kg m/s The negative sign indicates that the direction of the momentum is opposite to that before the collision. The change in momentum of the ball is given by: Δp = p2 - p1 Substituting the values, we get: Δp = (-4 kg m/s) - (4 kg m/s) = -8 kg m/s The negative sign indicates that the impulse experienced by the ball is in the opposite direction to its initial momentum, which is the direction of the wall. Therefore, the impulse experienced by the ball is 8 kg m/s. Therefore, the correct option is: 8kgm/s.